Concept Review CH. 15

Question 1

Suppose f is a continuous function defined on a rectangle R = [a,b] x [c,d]. Write an expression for double Riemann Sum of f. If f(x,y) >= 0, what does sum represent?
How is Midpoint Rule Used?

Answer 1

ΣΣ (f (xij, yij) Delta A.
If it is greater than 0 then it represents the volume underneath the rectangle.

Midpoint Rule just means f(xij, yij) is the centers of these rectangles.

Question 2

Definition of ∫∫ f(x,y) dA over R as a limit?

Answer 2

Lim n,m -> ∞ ΣΣ (f xij, yij) Delta A

Question 3

What is the Geometric Interpretation of ∫∫ f(x,y) dA over R if f(x,y) >= 0? What if f takes on both positive and negative values?

Answer 3

If f(x,y) >= 0, the integral represents the volume of the solid that lies above the rectangle R and below the surface z = f(x,y). If f takes on positive and negative values, then ∫∫ f(x,y) dA = V1 - V2. Where V1 is volume above R and below z = f(x,y). V2 is volume below R and above the surface.

Question 4

What is Fubini’s Theorem?

Answer 4

∫∫ f(x,y) dA = ∫a,b∫ c,d f(x,y) dxdy =

∫c,d∫ a,b f(x,y) dydx.

Question 5

How do you find the average value of f?

Answer 5

avg(f) = 1/A(R)  ∫∫ f(x,y) dA . 
A(R) = area of R, so if triangle, then find area of triangle you are iterating over.

Question 6

How do you define ∫∫D f(x,y) dA if D is a bounded region that is not a rectangle? What about for Triple Integrals?

Answer 6

Since D is bounded, then it can be enclosed in a rectangular region R. We define a new function F with domain R as 
F(x,y) = { f(x,y) if (x,y) is in D
               0      if (x,y) is in R, but not in D. } 
Then  ∫∫D f(x,y) dA  =  ∫∫R F(x,y) dA

Same process for triple integrals.

Question 7

What is the difference between a type 1 region and a type 2 region in R^2?

Answer 7

Type 1 means that the region D of ∫∫D f(x,y) dA lies between the graphs of two continuous functions of x, that is D = {x,y | a <= x <= b, g1(x) <= y <= g2(x)}.
Type 2 means it lies between the graphs of two functions of y, thus
D = {(x,y) | a <= y <= b, g1(y) <= x <= g2(y)}.

Question 8

What are the properties of Double Integrals?

Answer 8

They follow the same properties as Integrals. You know this and you can easily do double integrals.

Question 9

How do you change from rectangular coordinates to polar coordinates and why would you want to?

Answer 9

If we can more easily describe the region D in polar coordinates then we should make the change to polar coordinates.
D = { (r, θ) | a <= θ <= b, h1(θ) <= r <= h2(θ) }
x = r *cos(θ)
y = r *sin(θ)
dA = rdrdθ

Question 10

If a Plane region D has a density function p(x,y), then how do we find the mass, moments about the axes, and the center of mass?

Answer 10

mass = ∫∫D p(x,y) dA
Moments:
Mx = ∫∫D yp(x,y) dA
My = ∫∫D xp(x,y) dA

Center of Mass:
X = My/m
Y = Mx/m
(My/m, Mx/m)

Question 11

What is the equation for finding the surface area from a given function?

Answer 11

A(S) = ∫∫D √( (dx)^2 + (dy)^2 + 1 ) dA assuming dx and dy are the derivatives with respect to x and respect to y and assuming that they are continuous.

Question 12

What is the definition of a triple integral over a rectangular box B?

Answer 12

∫∫∫B f (x, y, z)dV = lim l,m, n -> ∞ ΣΣΣ f(xij, yij, zij) V.
V is the volume of each sub box.
Evaluate the same way as with double integrals.

Question 13

What is the difference between a Type 1 Region and Type 2 Region and Type 3 Region in R^3?

Answer 13

Type 1 if it lies between the graphs of two continuous functions of x and y, that is,
E = { (x,y,z)| (x,y) in D, u1(x,y) <= z <= u2(x,y) }
∫∫∫B f (x, y, z)dV = ∫∫ [ ∫u2(x,y) u1(x,y) f (x, y, z)dz ]dA

Type 2 if it lies between the graphs of two continuous functions of z and y, that is,
E = { (x,y,z)| (y, z) in D, u1(y, z) <= x <= u2(y,z) }
∫∫∫B f (x, y, z)dV = ∫∫ [ ∫u2(y, z) u1(y,z) f (x, y, z)dx ]dA

Type 3 if it lies between the graphs of two continuous functions of x and z, that is,
E = { (x,y,z)| (x,z) in D, u1(x,z) <= y <= u2(x,z) }
∫∫∫B f (x, y, z)dV = ∫∫ [ ∫u2(x,z) u1(x,z) f (x, y, z)dy ]dA

Question 14

What are the Mass, Center of Mass, Moments, and Center of Mass for Triple Integrals

Answer 14

m = ∫∫∫E p(x, y, z)dV for density function p.

Moments:
Myz = ∫∫∫E x *p(x, y, z)dV
Mxz = ∫∫∫E y *p(x, y, z)dV
Mxy = ∫∫∫E z *p(x, y, z)dV

Center of Mass:
(X, Y, Z) =( Myz/m, Mxz/m, Mxy/m)

Question 15

How to Change from Rectangular Coordinates to Cylindrical Coordinates in a triple Integral?

Answer 15

∫∫∫E f(x, y, z)dV = f (rcosθ, rsinθ, z) r dz dr dθ

E = {(r, θ, z) | a <= θ <= b, h1(θ) <= r <= h2(θ), u1( r cos(θ), r sin(θ) ) <= z <= u2( r cos(θ), r sin(θ) )

Question 16

How do we change from Rectangular Coordinates to Spherical Coordinates?

Answer 16

∫∫∫E f(x, y, z)dV = f (p sinɸ cosθ, p sinɸ sinθ, p cosɸ) p² sinɸ dp dθ dɸ
E = {(r, θ, ɸ) | a <= θ <= b, c <= ɸ <= d, g1( θ, ɸ ) <= p <=
g2( θ, ɸ )

Question 17

When should you change to cylindrical or spherical coordinates?

Answer 17

Regions with symmetry about the z axis are normally easier to describe with cylindrical coordinates. Regions that are symmetrical about the origin are easier to describe with spherical coordinates.

Question 18

If a transformation T is given by x = g(u,v) and y = h(u,v), what is the Jacobian of T?

Answer 18

d(x,y) | dx/du dx/dv |
__ = | dy/du dy/dv | = dx/du * dy/d/v - dxdv*dy/du
d(u,v)

Question 19

How do you change variables in a double integral? Triple Integral?

Answer 19

We can change from an integral in x and y to an integral in u and v by expressing x and y in terms of u and v and writing dA = | d(x,y) / d(u,v) | du dv
So find Jacobian then just that value du dv.
Thus:
∫∫R f(x,y) dA = ∫∫S f(x(u,v), y(u,v) ) | d(x,y) / d(u,v) | du dv

Triple Integral is the same way, except we then get to